# pH of 0.5M K2CO3

$\ce{H2CO3}$ ionises as \begin{align} \ce{H2CO3 + H2O &<=> H3O+ + HCO3-} & K_\mathrm a &= \pu{4.0*10^{-7}}\\ \ce{HCO3- + H2O &<=> H3O+ + CO3^2-} & K_\mathrm a &= \pu{5*10^{-11}}\\ \end{align}

The question is to find out the $\mathrm{pH}$ of $\pu{0.5 M}$ $\ce{K2CO3}$ solution.

Since $\ce{K2CO3}$ is a salt of weak acid and strong base hence its pH is given by $$\mathrm{pH} = \frac{\mathrm pK_\mathrm w + \mathrm pK_\mathrm a + \log c}{2}.$$

I am unsure of which $\mathrm pK_\mathrm a$ I should consider. Any help shall be appreciated.

• Consider obtaining $K_{\mathrm{b}}$ and solving equilibrium from the perspective of base.
– Zhe
Apr 17, 2017 at 21:55
• @Zhe we are given $K_a$ and not $K_b$. I think there will be anionic hydrolysis of the salt.
– Pink
Apr 17, 2017 at 22:03
• $\mathrm{K}_{b}$ of the conjugate base is $\mathrm{K}_{a}$ of the acid.
– Zhe
Apr 18, 2017 at 0:45
• @Zhe i know that $K_b$ of conjugate base is $K_a$ of acid but couldnot infer anything from this. can you please elaborate. Thanks.
– Pink
Apr 19, 2017 at 5:21
• $$\ce{CO3^{2-} + H2O <=> OH- + HCO3-},\ K_{\mathrm{b}} = ?$$
– Zhe
Apr 19, 2017 at 12:41

## 2 Answers

Write down the actual (equilibrium) reaction that develops between carbonate ion and water when one proton is exchanged (only full-fledged strong acids or bases might exchange a second proton to any significant extent). Which conjugate acid/base appears? $\ce {H2CO3/HCO3-}$ or $\ce {HCO3-/CO3^{2-}}$? When you see which one it is, use that dissociation constant.

The carbonate ion is the Conjugate base of the weak acid $\ce{HCO_3^-}\ (K_{a(HCO_3^-)}={5\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H_2CO_3}$ , so the solution of this problem as a solution of a monoprotic weak base: $\ce{CO_3^{-2} + H_2O <=> HCO_3^- + OH^-}$ $$\ce{K_{b(CO_3^{-2})}}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{K_{a(HCO_3^-)}}=\frac{10^{-14}}{5\times 10^{-11}}=\ce{2\times10^{-4}}$$ Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and little of the carbonate goes to bicarbonate,so$\ce{[CO_3^{-2}]}={[CO_3^{-2}]_0}=0.5$ ,so thus $$\dfrac{[OH^-]^2}{0.5}=K_{b(CO_3^{-2})}= \ce{2\times10^{-4}}$$ $\ce{[OH^-]}={10^{-2}}$ Which corresponds to $\ce{pOH = 2}$ or $\ce{pH = 12}$

• Re: The equilibrium expression must be solved as a quadratic yes and no. It depends on the precision of the answer. Little of the carbonate goes to bicarbonate, so to two significant figures you really don't need the quadratic solution. $$\sqrt{(2\times10^{-4})\times 0.5} = 0.010$$ so that solution is off by about 1 part in 50. Seems good enough.
– MaxW
Jul 30, 2018 at 23:04
• @MaxW I edit my answer your comment makes the calculation easy Jul 31, 2018 at 0:07
• Making the calculations easy is the point. In reality I'd be surprised if the "true" value was within 5% of the calculation. Also the "exact" solution, neglecting activities, is a cubic equation. So even the quadratic is an approximation.
– MaxW
Jul 31, 2018 at 0:12